This can be solved recursively.
{a1} -> a1
{a1, a2} -> {a1, a2}, {a2, a1}
{a1, a2, a3} -> {a1, a2, a3}, {a1, a3, a2}, {a3, a1, a2} , {a2, a1, a3}, {a2, a3, a1}, {a3, a2, a1}
So you can generate f(n) given f(n-1). pretty easy
Best paper awards at ICALP 2012
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The preliminary version of the detailed programme for ICALP 2012 is now
available here. While skimming through the programme, I learnt that the
best paper ...
38 minutes ago
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